Question

Suppose $z$ and $\omega$ are two complex numbers such that $\left|z\right|\le 1,\left|\omega \right|\le 1$, and $|z+i\omega |=|z-i\omega |=2$.

The complex number $z$ can be (where $\omega$ is cube root of unity)

• A. $1$ or $-i$
• B. $-1$
• C. $i$ or $-i$
• D. $\omega$ or ${\omega }^2$

Answer 1

Answer: (C)

$i$ or $-i$

$w$ is the cube root of unity.

$\therefore \left|w\right|=1\&w=\frac{-1+i\sqrt{3}}{2}$

$|z+iw|=|z-iw|=2$

$\Rightarrow \left|\frac{z+iw}{z-iw}\right|=1$

$\Rightarrow \frac{z+iw}{z-iw}=\cos A+i\sin A$

$\Rightarrow \frac{z+iw}{z-iw}=\cos A+i\sin A$

$\Rightarrow z=-iw\left(\frac{1+\cos A+i\sin A}{1-\cos A-i\sin A}\right)$

$\Rightarrow z=-iw\left(\frac{2{\cos }^2\frac{A}{2}+2i\cos \frac{A}{2}\sin \frac{A}{2}}{2{\sin }^2\frac{A}{2}-2i\cos \frac{A}{2}\sin \frac{A}{2}}\right)=-iw\cot \frac{A}{2}\left(\frac{\cos \frac{A}{2}+i\sin \frac{A}{2}}{\sin \frac{A}{2}-i\cos \frac{A}{2}}\right)$

$\Rightarrow z=w\cot \frac{A}{2}$

$|z+iw|=2$

$\Rightarrow |w\cot \frac{A}{2}+iw|=2$

$\Rightarrow {\left|w\right|}^2(1+{\cot }^2\frac{A}{2})=4$

$\Rightarrow {\sin }^2\frac{A}{2}=\frac{1}{4}$

$\therefore A=\frac{\pi }{3}$

$\because z=w\cot \frac{A}{2}$

$\therefore z=w\cot \frac{\pi }{6}=\left(\frac{-1+i\sqrt{3}}{2}\right)\sqrt{3}$